IJMMS 25:4 (2001) 217–229
PII. S0161171201004707
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
ON THE STABILITY OF THE QUADRATIC
MAPPING IN NORMED SPACES
GWANG HUI KIM
(Received 3 December 1999 and in revised form 28 February 2000)
Abstract. The Hyers-Ulam stability, the Hyers-Ulam-Rassias stability, and also the stability in the spirit of Gǎvruţa for each of the following quadratic functional equations
f (x + y) + f (x − y) = 2f (x) + 2f (y), f (x + y + z) + f (x − y) + f (y − z) + f (z − x) =
3f (x)+3f (y)+3f (z), f (x +y +z)+f (x)+f (y)+f (z) = f (x +y)+f (y +z)+f (z +x)
are investigated.
2000 Mathematics Subject Classification. Primary 39B52, 39B72, 39B82.
1. Introduction. The stability problem of functional equations was originally raised
by Ulam [8] in 1940. He posed the following problem: under what conditions does
there exist an additive mapping near an approximately additive mapping? In 1941,
this problem was solved by Hyers [1] in the case of Banach space. Thereafter, we
call that type the Hyers-Ulam stability. In 1978, Rassias [6] extended the Hyers-Ulam
stability by considering variables. In 1994, it also has been generalized to the function
case by Gǎvruţa [3]. Throughout this paper, let X and Y be a real normed space and
a real Banach space, respectively. Also R and N stand for the set of all real numbers
and natural numbers, respectively.
The quadratic function f (x) = x 2 is a solution of each of the following functional
equations
f (x + y) + f (x − y) = 2f (x) + 2f (y),
(1.1)
f (x + y + z) + f (x − y) + f (y − z)+f (z − x) = 3f (x) + 3f (y) + 3f (z),
(1.2)
f (x + y + z) + f (x) + f (y) + f (z) = f (x + y) + f (y + z) + f (z + x).
(1.3)
So, it is natural that each equation is called a quadratic functional equation. In particular, every solution of the “original” quadratic functional equation (1.1) is said to be
a quadratic function.
For the quadratic functional equation some results are contained in [1, 2, 4, 7].
Skof [7] and Cholewa [1] proved a Hyers-Ulam stability theorem of the quadratic functional equation (1.1) in different domains. Czerwik proved in [2] a Hyers-Ulam-Rassias
stability for (1.1) which contains the following theorem as a particular case.
Theorem 1.1. Let δ ≥ 0 be fixed. If f : X → Y satisfies the inequality
f (x + y) + f (x − y) − 2f (x) − 2f (y) ≤ δ
∀x ∈ X,
(1.4)
218
GWANG HUI KIM
then there exists a unique quadratic mapping g : X → Y such that
g(x) − f (x) ≤ δ
2
∀x ∈ X.
(1.5)
If, moreover, f is measurable or f (tx) is continuous in t for each fixed x ∈ X, then
g(tx) = t 2 g(x) for all x ∈ X and t ∈ R.
The stability in the sense of Rassias for (1.1) on a restricted domain and for (1.3) is
proved by Jung [4]. In Section 2, the stability in the spirit of Gǎvruţa of (1.1) (more generally, modified Hyers-Ulam-Rassias stability) is investigated. In Section 3, the HyersUlam stability, the Hyers-Ulam-Rassias stability, and the stability in the spirit of
Gǎvruţa of (1.2) are investigated. In Section 4, the stability in the spirit of Gǎvruţa
of (1.3) under the approximately even (or odd) condition is treated.
2. Stability in the spirit of Gǎvruţa of (1.1). The stability of the quadratic functional equation (1.1) is proved under the spirit of Gǎvruţa. Let mappings ϕ and
Φ : X × X → [0, ∞) satisfy the inequality
Φ(x, y) =
∞
1
1
ϕ(0, 0) +
ϕ 2k x, 2k y < ∞
k+1
6
4
k=0
∀x, y ∈ X.
(2.1)
By using an idea in Gǎvruţa [3] we can prove the following results.
Lemma 2.1. Assume that f : X → Y satisfies the inequality
f (x + y) + f (x − y) − 2f (x) − 2f (y) ≤ ϕ(x, y)
∀x, y ∈ X.
(2.2)
Then, for all x ∈ X and n ∈ N,
n−1
n−1
n 1
f 2 x − 4n f (x) ≤
4k ϕ(0, 0) +
4k ϕ 2n−1−k x, 2n−1−k x .
2
k=0
k=0
(2.3)
Proof. Put x = y = 0 in (2.2) and conclude that
f (0) ≤
1
ϕ(0, 0).
2
(2.4)
For x = y the inequality (2.2) again implies
f (2x) − 4f (x) ≤ 1 ϕ(0, 0) + ϕ(x, x)
2
∀x ∈ X
(2.5)
which proves the inequality (2.3) for n = 1. For the induction, we assume that (2.3)
ON THE STABILITY OF THE QUADRATIC MAPPING IN NORMED SPACES
219
holds for some n ∈ N. Then, for any x ∈ X, we have, for n + 1,
n+1 f 2
x − 4n+1 f (x) ≤ f 2 · 2n x − 4f 2n x + 4f 2n x − 4n f (x)
≤
=
1
ϕ(0, 0) + ϕ 2n x, 2n x
2


n−1
n−1
n−1−k
k1
k
n−1−k

+4
4 ϕ(0, 0) +
4 ϕ 2
x, 2
x 
2
k=0
k=0
(2.6)
n
n
1
4k ϕ(0, 0) +
4k ϕ 2n−k x, 2n−k x
2
k=0
k=0
which proves the inequality (2.3) for all natural n.
Theorem 2.2. If f : X → Y satisfies the inequality (2.2), then there exists a unique
quadratic mapping g : X → Y such that
g(x) − f (x) ≤ Φ(x, x) ∀ x ∈ X.
(2.7)
If, moreover, f is measurable or f (tx) is continuous in t for each fixed x ∈ X, then
g(tx) = t 2 g(x) for all x ∈ X and t ∈ R.
Proof. For any x ∈ X and for every positive integer n ∈ N, we define
gn (x) = 4−n f 2n x .
(2.8)
From (2.3), we have, for n > m,
gn (x) − gm (x) = 4−n f 2n−m x · 2m x − 4n−m f 2m x 

n−m−1
n−m−1
n−1−k
−n 
k1
k
n−1−k
≤4
4 ϕ(0, 0) +
4 ϕ 2
x, 2
x 
2
k=0
k=0
=
n−m−1
k=0
=
n−m−1
1
4−n+k ϕ(0, 0) +
4−n+k ϕ 2n−1−k x, 2n−1−k x
2
k=0
(2.9)
n
n
1
4−k ϕ(0, 0) +
4−k ϕ 2k−1 x, 2k−1 x .
2
k=m+1
k=m+1
By (2.1), since the right-hand side of the above inequality tends to zero as m tends
to infinity, the sequence {gn (x)} is a Cauchy sequence for all x ∈ X. Since Y is a
Banach space, we define a function g : X → Y by
g(x) = lim gn (x) ∀x ∈ X.
n→∞
(2.10)
From the inequality (2.2), it follows that
gn (x + y) + gn (x − y) − 2gn (x) − 2gn (y)
= 4−n f 2n x + 2n y + f 2n x − 2n y − 2f 2n x − 2f 2n y ≤4
1
4n+1
n
n
ϕ 2 x, 2 y
(2.11)
220
GWANG HUI KIM
for all x, y ∈ X and n ∈ N. Therefore, by letting n → ∞ in the last inequality we obtain
(1.1) from (2.1). Moreover, from Lemma 2.1, for all x ∈ X and n ∈ N we have the
inequality:
gn (x) − f (x) = 4−n f 2n x − 4n f (x)


n−1
n−1
n−1−k
−n 
k1
k
n−1−k
≤4
4 ϕ(0, 0) +
4 ϕ 2
x, 2
x 
2
k=0
k=0
=
n−1
(2.12)
n−1
1
4−n+k ϕ(0, 0) +
4−n+k ϕ 2n−1−k x, 2n−1−k x .
2
k=0
k=0
Hence from (2.1) we see that (2.7) holds true.
If h : X → Y is another function which satisfies (1.1) and (2.7), since g(0) = 0 = h(0),
then by (1.1) we have
g 2n x = 4n g(x),
h 2n x = 4n h(x)
(2.13)
for all x ∈ X and n ∈ N. Hence, by (2.7) it follows that
g(x) − h(x) = 4−n g 2n x − h 2n x ≤ 4−n g 2n x − f 2n x + f 2n x − h 2n x 2Φ 2n x, 2n x
≤
4n
(2.14)
for all x ∈ X and n ∈ N. By letting n → ∞ in the preceding inequality, we immediately
see the uniqueness of g. The proof of the last assertion in the theorem goes through
in the same way as that of Theorem 1.1 (see [2, Theorem 1]).
Note. The last assertion in all results of this paper goes through in the same way
as that of Theorem 1.1 (see [2, Theorem 1]).
The following corollary is the Hyers-Ulam stability of quadratic functional equation (1.1) which is the result of Skof [7] and Cholewa [1]. Applying Theorem 2.2 with
ϕ(x, y) = δ, we get the following corollary.
Corollary 2.3. If f : X → Y satisfies the inequality
f (x + y) + f (x − y) − 2f (x) − 2f (y) ≤ δ,
(2.15)
then there exists a quadratic mapping g : X → Y satisfying (1.1) and such that
g(x) − f (x) ≤ δ
2
∀ x ∈ X.
(2.16)
If, moreover, f is measurable or f (tx) is continuous in t for each fixed x ∈ X, then
g(tx) = t 2 g(x) for all x ∈ X and t ∈ R.
The following theorem is the Hyers-Ulam-Rassias stability of quadratic functional
equation (1.1).
ON THE STABILITY OF THE QUADRATIC MAPPING IN NORMED SPACES
221
Theorem 2.4. Let X be a normed space and Y a Banach space and let ξ, θ ≥ 0 and
p < 2 be given real numbers. Let f : X → Y be a function satisfying the inequality
f (x + y) + f (x − y) − 2f (x) − 2f (y) ≤ ξ + θ xp + yp
∀x, y ∈ X.
(2.17)
Then there exists exactly one quadratic mapping g : X → Y such that
g(x) − f (x) ≤ 1 ξ + 2 4 − 2p −1 θxp ,
2
x ∈ X.
(2.18)
If, moreover, f is measurable or f (tx) is continuous in t for each fixed x ∈ X, then
g(tx) = t 2 g(x) for all x ∈ X and t ∈ R.
Theorem 2.4 is easily proved using Theorem 2.2.
Corollary 2.5 [2, Theorem 1]. Let X be a normed space and Y a Banach space and
let ξ, θ ≥ 0 and p < 2 be given real numbers. Let f : X → Y be a function satisfying the
inequality
f (x +y)+f (x −y)−2f (x)−2f (y) ≤ ξ +θ xp +yp
∀x, y ∈ X \{0}. (2.19)
Then there exists exactly one quadratic mapping g : X → Y such that
g(x) − f (x) ≤ ξ + f (0) + 2 4 − 2p −1 θxp ,
3
x ∈ X \ {0}.
(2.20)
If, moreover, f is measurable or f (tx) is continuous in t for each fixed x ∈ X, then
g(tx) = t 2 g(x) for all x ∈ X and t ∈ R.
Proof. For x = y ∈ X \ {0} the inequality (2.2) in Lemma 2.1 implies
f (2x) − 4f (x) ≤ f (0) + ϕ(x, x)
∀ x ∈ X \ {0},
(2.21)
then we have
n−1
n n−1
f 2 x − 4n f (x) ≤
4k f (0) +
4k ϕ 2n−1−k x, 2n−1−k x
k=0
∀ x ∈ X \ {0}.
k=0
(2.22)
By applying Theorem 2.2 with ϕ(x, y) = θ(xp + yp ) for p < 2, the proof of the
corollary is complete.
3. Three types stability of (1.2). In this section, we investigate the Hyers-Ulam stability, Hyers-Ulam-Rassias stability, and the stability in the spirit of Gǎvruţa for (1.2).
Lemma 3.1. Assume that f : X → Y satisfies the inequality
f (x + y + z) + f (x − y) + f (y − z) + f (z − x) − 3f (x) − 3f (y) − 3f (z) ≤ δ
(3.1)
for all x, y, z ∈ X and δ ≥ 0. Then for x ∈ X and n ∈ N,
n
n f 3 x − 32n f (x) ≤ 8 δ
32(k−1) .
5 k=1
(3.2)
222
GWANG HUI KIM
Proof. Put x = y = z = 0 in (3.1) and conclude that
f (0) ≤
δ
.
5
(3.3)
For x = y = z, the inequality (3.1) again implies
f (3x) − 32 f (x) ≤ δ + 3f (0) ≤ 8 δ
5
(3.4)
which proves the inequality (3.2) for n = 1. For the induction, we assume that (3.2)
holds for some n ∈ N. Then, for any x ∈ X, by (3.4) we have, for n + 1,
n+1 f 3
x − 32(n+1) f (x)
≤ f 3 · 3n x − 32 f 3n x + 32 f 3n x − 32n f (x)
(3.5)
n
n+1
n+1
8
8
8 2(k−1)
2 8
2(k−1)
2(k−1)
≤ δ+3
δ
3
3
3
= δ 1+
= δ
5
5 k=1
5
5 k=1
k=2
which proves the inequality (3.2) for all natural n.
Theorem 3.2. Assume that a mapping f : X → Y satisfies the inequality (3.1). Then
there exists a unique quadratic mapping g : X → Y satisfying (1.2) and the inequality
g(x) − f (x) ≤ δ
5
∀ x ∈ X.
(3.6)
If, moreover, f is measurable or f (tx) is continuous in t for each fixed x ∈ X, then
g(tx) = t 2 g(x) for all x ∈ X and t ∈ R.
Proof. For any x ∈ X and for every positive integer n ∈ N, we define
gn (x) = 3−2n f 3n x .
(3.7)
From (3.4), we have
gn+1 (x) − gn (x) = 3−2(n+1) f 3 · 3n x − 32 f 3n x ≤ 3−2(n+1) 8 δ
5
(3.8)
for all n > N and for all x ∈ X. Hence we have, for n ≥ m,
n−1
n−1
gj+1 (x) − gj (x) ≤ 8 δ
gn (x) − gm (x) ≤
3−2(j+1) → 0
5
j=m
j=m
(3.9)
for all n > N and for all x ∈ X as m → ∞. We see that the sequence {gn (x)} is a
Cauchy sequence. Hence, we can define a function g : X → Y by
g(x) = lim gn (x) ∀x ∈ X.
n→∞
(3.10)
Then for all x, y, z ∈ X and n ∈ N, we have, from (3.1),
gn (x + y + z) + gn (x − y) + gn (y − z) + gn (z − x) − 3gn (x) − 3gn (y) − 3gn (z)
= 3−2n f 3n (x + y + z) + f 3n (x − y) + f 3n (y − z) + f 3n (z − x)
δ
− 3f 3n x − 3f 3n y − 3f 3n z ≤ 2n .
3
(3.11)
ON THE STABILITY OF THE QUADRATIC MAPPING IN NORMED SPACES
223
Letting n tend to infinity, we obtain (1.2). Moreover, from Lemma 3.1, we have
gn (x) − f (x) = 3−2n f 3n x − 32n f (x)
n
n
8 2(k−1) 8 −2k
3
= δ
3
≤ 3−2n δ
5 k=1
5 k=1
(3.12)
for all x, y ∈ X and n ∈ N. By letting n tend to infinity, we obtain the inequality (3.6).
The proof of the uniqueness is the same way as that of Theorem 2.2 by applying
g(3n x) = 32n g(x) and h(3n x) = 32n h(x). Hence, the proof is complete.
Let the mappings ϕ and Φ : X × X × X → [0, ∞) satisfy the inequality
Φ(x, y, z) =
∞
3
ϕ0 +
3−3k ϕ 3k−1 x, 3k−1 y, 3k−1 z < ∞.
130
k=1
(3.13)
For simplicity of calcuation in this section, we use the notation ϕx = ϕ(x, x, x).
Lemma 3.3. Assume that f : X → Y satisfies the inequality
f (x + y + z) + f (x − y) + f (y − z) + f (z − x) − 3f (x) − 3f (y) − 3f (z) ≤ ϕ(x, y, z)
(3.14)
for all x, y, z ∈ X. It then holds that for all x ∈ X and for all n ∈ N,
n
n
n f 3 x − 33n f (x) ≤ 3 ϕ0
33(k−1) +
33(n−k) ϕ3k−1 x .
5
k=1
k=1
(3.15)
Proof. Put x = y = z = 0 in (3.14) and conclude that
f (0) ≤
1
ϕ0 .
5
(3.16)
For x = y = z, the inequality (3.14) again implies
f (3x) − 33 f (x) ≤ 3f (0) + ϕx ≤ 3 ϕ0 + ϕx
5
(3.17)
which proves the inequality (3.15) for n = 1. For the induction, we assume that (3.15)
holds for some n ∈ N and for all x ∈ X. Then by (3.17) we have, for n + 1,
n+1 f 3
x − 33(n+1) f (x) ≤ f 3 · 3n x − 33 f 3n x + 33 f 3n x − 33n f (x)
n
n
3
3 3
3(k−1)
3(n−k)
≤ ϕ0 + ϕ3n x + 3
ϕ0
3
+
3
ϕ3k−1 x
5
5
k=1
k=1
=
n+1
n+1
3
ϕ0
33(k−1) +
33(n−k) ϕ3k−1 x
5
k=1
k=1
(3.18)
which proves the inequality (3.15) for all natural n.
Theorem 3.4. Assume that a mapping f : X → Y satisfies the equality (3.14). Then
there exists a unique quadratic mapping g : X → Y that satisfies (1.2) and the inequality
g(x) − f (x) ≤ Φx
∀ x ∈ X.
(3.19)
224
GWANG HUI KIM
If, moreover, f is measurable or f (tx) is continuous in t for each fixed x ∈ X, then
g(tx) = t 2 g(x) for all x ∈ X and t ∈ R.
Proof. For every positive integer n, we define
gn (x) = 3−3n f 3n x
∀ x ∈ X.
(3.20)
By (3.17), we have
gn+1 (x) − gn (x) = 3−3(n+1) f 3 · 3n x − 33 f 3n x 3
≤ 3−3(n+1) ϕ0 + 3−3(n+1) ϕ3n x
5
(3.21)
for all n ∈ N and for all x ∈ X. Hence by (3.21) we have, for n ≥ m,
n−1
gj+1 (x) − gj (x)
gn (x) − gm (x) ≤
j=m
n−1
n−1
3
3−3(j+1) +
3−3(j+1) ϕ3j x
≤ ϕ0
5
j=m
j=m
(3.22)
for all x ∈ X. By (3.13) since the right-hand side of the preceding inequality tends
to zero as m tends to infinity, we see that the sequence {gn (x)} is a Cauchy sequence.
Hence we can define a function g : X → Y by
g(x) = lim gn (x) ∀x ∈ X.
n→∞
(3.23)
Then for all x, y, z ∈ X and n ∈ N, we have
gn (x + y + z) + gn (x − y) + gn (y − z) + gn (z − x) − 3gn (x) − 3gn (y) − 3gn (z)
= 3−3n f 3n (x + y + z) + f 3n (x − y) + f 3n (y − z) + f 3n (z − x)
ϕ 3n x, 3n y, 3n z
− 3f 3n x − 3f 3n y − 3f 3n z ≤
.
33n
(3.24)
Letting n tend to infinity, we obtain (1.2). Moreover, from Lemma 3.3, we have
gn (x) − f (x) = 3−3n f 3n x − 33n f (x)
n
n
−3n 3
3(k−1)
3(n−k)
ϕ0
3
+
3
ϕ3k−1 x
≤3
5
k=1
k=1
n
n
3
ϕ0
3−3(n−k+1) +
3−3k ϕ3k−1 x
5
k=1
k=1
n
1
3
1 − 3n ϕ0 +
=
3−3k ϕ3k−1 x
130
3
k=1
=
(3.25)
for all x ∈ X and n ∈ N. From (3.13) as n tend to infinity we obtain the inequality (3.19).
The proof of the uniqueness is the same way as that of Theorem 2.2 by applying
g(3n x) = 32n g(x) and h(3n x) = 32n h(x). Hence, the proof is complete.
ON THE STABILITY OF THE QUADRATIC MAPPING IN NORMED SPACES
225
The following theorem is the Hyers-Ulam-Rassias stability of quadratic functional
equation (1.2).
Theorem 3.5. Assume that a mapping f : X → Y satisfies the equality
f (x+y+z)+f (x−y)+f (y−z)+f (z−x) − 3f (x)−3f (y)−3f (z)
≤ θ xp +yp +zp
(3.26)
for all x, y, z ∈ X and p < 3. Then there exists a unique quadratic mapping g : X → Y
satisfies (1.2) and the inequality
g(x) − f (x) ≤ 333 − 3p −1 θxp
∀ x ∈ X.
(3.27)
If, moreover, f is measurable or f (tx) is continuous in t for each fixed x ∈ X, then
g(tx) = t 2 g(x) for all x ∈ X and t ∈ R.
Proof. Apply Theorem 3.2, with ϕ(x, y, z) = θ(xp + yp + zp ), and (3.13).
Then we obtain
Φ(x, x, x) =
∞
p p p 3−3k θ 3k−1 x + 3k−1 x + 3k−1 x k=1
= 3θx
p
∞
−3k (k−1)p
3
3
3
= 3 3 −3
p −1
(3.28)
p
θx .
k=1
4. Stability in the spirit of Gǎvruţa of (1.3). In this section, the stability of another
quadratic (1.3) is investigated under the spirit of Gǎvruţa. The Hyers-Ulam-Rassias
stability of (1.3) can be found in [4].
Let the mappings ϕ and Φ : X × X × X → [0, ∞) satisfy the inequality
Φ(x, y, z) = 2ϕ(0, 0, 0) +
+
∞
2k + 1 k−1
ϕ 2 x, 2k−1 y, −2k−1 z
2k+1
2
k=0
∞
2k − 1 ϕ − 2k−1 x, −2k−1 y, 2k−1 z < ∞
2k+1
2
k=0
(4.1)
∀x ∈ X.
For simplicity of calculation in this section, we use the notation ϕx = ϕ(x, x, −x).
Lemma 4.1. Assume that a mapping f : X → Y satisfies the following inequality:
f (x +y +z)+f (x)+f (y)+f (z)−f (x +y)−f (y +z)−f (z+x) ≤ ϕ(x, y, z) (4.2)
for all x, y, z ∈ X. It then holds that
n
n
f (x) − 2 + 1 f 2n x + 2 − 1 f − 2n x 22n+1
22n+1
n
2k + 1
2k − 1
1
≤
ϕ
+
ϕ
+
ϕ
k−1
k−1
0
2k−1
22k+1 2 x 22k+1 −2 x
k=1
for all x, y, z ∈ X and n ∈ N.
(4.3)
226
GWANG HUI KIM
Proof. Put x = y = z = 0 in (4.2) and conclude that f (0) ≤ ϕ0 . And also putting
x = y = −z in (4.2) yields
3f (x) + f (−x) − f (2x) ≤ 2ϕ0 + ϕx .
(4.4)
Substitute −x for x in (4.4), we obtain
3f (−x) + f (x) − f (−2x) ≤ 2ϕ0 + ϕ−x .
We use induction on n to prove our lemma. By (4.4) and (4.5), we have
f (x) − 3 f (2x) + 1 f (−2x)
8
8
1
3
≤ 3f (x) + f (−x) − f (2x) + 3f (−x) + f (x) − f (−2x)
8
8
1
3
3
1
≤
2ϕ0 + ϕx + 2ϕ0 + ϕ−x = ϕ0 + ϕx + ϕ−x
8
8
8
8
(4.5)
(4.6)
which proves the validity of the inequality (4.3) for the case n = 1. Now assume that
the inequality (4.3) holds true for some n ∈ N. By using (4.4) and (4.5), we have the
following relation:
n+1
+ 1 n+1 2n+1 − 1 n+1
f (x)− 2
f
2
x
+
f
−
2
x
22n+3
22n+3
n
n
2 +1 n 2 −1 n
≤
f (x) − 22n+1 f 2 x + 22n+1 f − 2 x 2n+1 + 1 3f 2n x + f − 2n x − f 2n+1 x 2n+3
2
2n+1 − 1 + 2n+3 3f − 2n x + f 2n x − f − 2n+1 x 2
n 2k + 1
2k − 1
1
≤
ϕ
+
ϕ
ϕ
k−1 x +
k−1 x
0
2
−2
2k−1
22k+1
22k+1
k=1
+
(4.7)
2n+1 − 1 2n+1 + 1 2ϕ0 + ϕ2n x + 2n+3 2ϕ0 + ϕ−2n x
2n+3
2
2
n+1
1
2k + 1
2k − 1
=
ϕ0 + 2k+1 ϕ2k−1 x + 2k+1 ϕ−2k−1 x
k−1
2
2
2
k=1
+
which proves the inequality (4.3) for n + 1.
Theorem 4.2. Assume that a mapping f : X → Y satisfies the inequalities (4.2) and
f (x) − f (−x) ≤ θ
(4.8)
for θ ≥ 0 and for all x, y, z ∈ X. Then there exists a unique quadratic mapping g : X → Y
satisfying (1.3) such that
g(x) − f (x) ≤ Φ(x) ∀ x ∈ X.
(4.9)
If, moreover, f is measurable or f (tx) is continuous in t for each fixed x ∈ X, then
g(tx) = t 2 g(x) for all x ∈ X and t ∈ R.
ON THE STABILITY OF THE QUADRATIC MAPPING IN NORMED SPACES
227
Proof. For any x ∈ X and for every positive integer n, we define
gn (x) = 2−2n f 2n x .
(4.10)
From (4.4) and (4.8), we have
gn+1 (x) − gn (x)
= 2−2(n+1) f 2 · 2n x − 22 f 2n x (4.11)
≤ 2−2(n+1) 3f 2n x + f − 2n x − f 2 · 2n x + f 2n x − f − 2n x ≤ 2−2(n+1) 2ϕ0 + ϕ2n x + θ
for all x ∈ X and for all n ∈ N. Therefore we have, for n ≥ m,
n−1
gn (x) − gm (x) ≤
gj+1 (x) − gj (x)
j=m
≤
n−1
j=m
2−2(j+1) 2ϕ0 + ϕ2j x + θ
(4.12)
n−1
n−1
2−2(j+1) +
2−2(j+1) ϕ2j x
≤ 2ϕ0 + θ
j=m
j=m
for all x ∈ X. By (4.1), since the right-hand side of the inequality (4.12) tends to zero
as m tends to infinity, the sequence {gn (x)} is a Cauchy sequence for all x ∈ X, and
hence we define a function g : X → Y by
g(x) = lim gn (x) ∀x ∈ X.
n→∞
The inequality (4.2) implies that
n
gn 2 (x + y + z) + gn 2n x + gn 2n y + gn 2n z − gn 2n (x + y)
− gn 2n (y + z) − gn 2n (z + x) ≤ 2−2n ϕ 2n x, 2n y, 2n z
(4.13)
(4.14)
for all x, y, z ∈ X and n ∈ N. Letting n tend to infinity in the last inequality, then by
(4.1) we obtain (1.3). Analogously, by (4.8), we can see that g is even. By substituting
−y for z in (1.3) and by taking account of g(0) = 0, we see that g as an even solution
of (1.3) is quadratic. From (4.3) and (4.8), we have
n
n
f (x) − gn (x) ≤ f (x) − 2 + 1 f 2n x + 2 − 1 f − 2n x 22n+1
22n+1
−2n n 2n + 1 n 2n − 1 n
+
2
f
2
x
−
f
2
x
+
f
−
2
x
(4.15)
22n+1
22n+1
n
2k + 1
2k − 1
1
2n − 1
≤
ϕ0 + 2k+1 ϕ2k−1 x + 2k+1 ϕ−2k−1 x + 2n+1 θ
k−1
2
2
2
2
k=1
for all x ∈ X and for all n ∈ N.
According to (4.1) and (4.15), the inequality (4.9) holds true.
The proof of the uniqueness is similar to that of Theorem 2.2 by applying g(2n x) =
4n g(x) and h(2n x) = 4n h(x). Hence, the proof is complete.
228
GWANG HUI KIM
Corollaries 4.3 and 4.5 are the Hyers-Ulam stability of quadratic functional equation (1.3) under the approximately even or odd condition which is the result of Jung
[4]. For the proof, apply Theorems 3.5 and 4.2 with ϕ(x, y, z) = δ.
Corollary 4.3. Assume a mapping f : X → Y satisfies the inequality
f (x + y + z) + f (x) + f (y) + f (z) − f (x + y) − f (y + z) − f (z + x) ≤ δ,
f (x) − f (−x) ≤ θ,
(4.16)
for some δ, θ ≥ 0 and for all x, y, z ∈ X. Then there exists a unique quadratic mapping
g : X → Y satisfying (1.3) and the inequality
g(x) − f (x) ≤ 4δ
∀x ∈ X.
(4.17)
If, moreover, f is measurable or f (tx) is continuous in t for each fixed x ∈ X, then
g(tx) = t 2 g(x) for all x ∈ X and t ∈ R.
Theorem 4.4. Assume that a mapping f : X → Y satisfies the inequalities (4.2) and
f (x) + f (−x) ≤ θ
(4.18)
for θ ≥ 0 and for all x, y, z ∈ X. Then there exists a unique additive mapping g : X → Y
satisfying (1.1) and
g(x) − f (x) ≤ Φ(x) ∀ x ∈ X.
(4.19)
If, moreover, f is measurable or f (tx) is continuous in t for each fixed x ∈ X, then
g(tx) = t 2 g(x) for all x ∈ X and t ∈ R.
Proof. For any x ∈ X and for every positive integer n, we define
gn (x) = 2−n f 2n x .
(4.20)
From (4.4) and (4.18), we have
gn+1 (x) − gn (x)
= 2−(n+1) f 2 · 2n x − 2f 2n x ≤ 2−(n+1) 3f 2n x + f − 2n x − f 2 · 2n x + f 2n x + f − 2n x ≤ 2−(n+1) 2ϕ0 + ϕ2n x + θ
(4.21)
for all x ∈ X and for all n ∈ N. Therefore we have, for n ≥ m,
n−1
gn (x) − gm (x) ≤
gj+1 (x) − gj (x)
j=m
≤
n−1
j=m
2−(j+1) 2ϕ0 + ϕ2j x + θ
n−1
n−1
2−(j+1) +
2−(j+1) ϕ2j x
≤ 2ϕ0 + θ
j=m
j=m
(4.22)
ON THE STABILITY OF THE QUADRATIC MAPPING IN NORMED SPACES
229
for all x ∈ X. By definition (4.1) of Φ, since the right-hand side of the inequality (4.22)
tends to zero as m tends to infinity, the sequence {gn (x)} is a Cauchy sequence for
all x ∈ X, and hence we define a function g : X → Y by
g(x) = lim gn (x) ∀x ∈ X.
n→∞
(4.23)
Similarly, as in the proof of Theorem 4.2, due to (4.18), we see that the mapping g
satisfies (1.3) and is odd. By putting z = −y in (1.3) and considering the oddness of g
and letting u = x + y, v = x − y, we get
u+v
2g
= g(u) + g(v).
(4.24)
2
According to [5], since g(0) = 0, the mapping g is additive. Similarly—as in the proof
of (4.9) of Theorem 4.2—from Lemma 4.1, (4.18), and (4.1), we directly see that (4.19)
holds true.
The proof of the uniqueness is similar to that of Theorem 2.2 by applying g(2n x) =
2n g(x) and h(2n x) = 2n h(x). Hence, the proof is complete.
Corollary 4.5. Assume a mapping f : X → Y satisfies the inequality
f (x + y + z) + f (x) + f (y) + f (z) − f (x + y) − f (y + z) − f (z + x) ≤ δ,
f (x) + f (−x) ≤ θ,
(4.25)
for some δ, θ ≥ 0 and for all x, y, z ∈ X. Then there exists a unique additive mapping
g : X → Y satisfying the inequality
g(x) − f (x) ≤ 4δ ∀x ∈ X.
(4.26)
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Gwang Hui Kim: Department of Mathematics, Kangnam University, Suwon, 449-702,
Korea
E-mail address: ghkim@kns.kangnam.ac.kr